3.1.12 \(\int \frac {\cos ^2(a+b x^2)}{x} \, dx\) [12]

Optimal. Leaf size=37 \[ \frac {1}{4} \cos (2 a) \text {CosIntegral}\left (2 b x^2\right )+\frac {\log (x)}{2}-\frac {1}{4} \sin (2 a) \text {Si}\left (2 b x^2\right ) \]

[Out]

1/4*Ci(2*b*x^2)*cos(2*a)+1/2*ln(x)-1/4*Si(2*b*x^2)*sin(2*a)

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Rubi [A]
time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3485, 3459, 3457, 3456} \begin {gather*} \frac {1}{4} \cos (2 a) \text {CosIntegral}\left (2 b x^2\right )-\frac {1}{4} \sin (2 a) \text {Si}\left (2 b x^2\right )+\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^2/x,x]

[Out]

(Cos[2*a]*CosIntegral[2*b*x^2])/4 + Log[x]/2 - (Sin[2*a]*SinIntegral[2*b*x^2])/4

Rule 3456

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3457

Int[Cos[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[CosIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3459

Int[Cos[(c_) + (d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Dist[Cos[c], Int[Cos[d*x^n]/x, x], x] - Dist[Sin[c], Int[Si
n[d*x^n]/x, x], x] /; FreeQ[{c, d, n}, x]

Rule 3485

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Cos[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^2\left (a+b x^2\right )}{x} \, dx &=\int \left (\frac {1}{2 x}+\frac {\cos \left (2 a+2 b x^2\right )}{2 x}\right ) \, dx\\ &=\frac {\log (x)}{2}+\frac {1}{2} \int \frac {\cos \left (2 a+2 b x^2\right )}{x} \, dx\\ &=\frac {\log (x)}{2}+\frac {1}{2} \cos (2 a) \int \frac {\cos \left (2 b x^2\right )}{x} \, dx-\frac {1}{2} \sin (2 a) \int \frac {\sin \left (2 b x^2\right )}{x} \, dx\\ &=\frac {1}{4} \cos (2 a) \text {Ci}\left (2 b x^2\right )+\frac {\log (x)}{2}-\frac {1}{4} \sin (2 a) \text {Si}\left (2 b x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 34, normalized size = 0.92 \begin {gather*} \frac {1}{4} \left (\cos (2 a) \text {CosIntegral}\left (2 b x^2\right )+2 \log (x)-\sin (2 a) \text {Si}\left (2 b x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^2/x,x]

[Out]

(Cos[2*a]*CosIntegral[2*b*x^2] + 2*Log[x] - Sin[2*a]*SinIntegral[2*b*x^2])/4

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 68, normalized size = 1.84

method	result	size
risch	\(\frac {\ln \left (x \right )}{2}+\frac {i {\mathrm e}^{-2 i a} \pi \,\mathrm {csgn}\left (b \,x^{2}\right )}{8}-\frac {i {\mathrm e}^{-2 i a} \sinIntegral \left (2 b \,x^{2}\right )}{4}-\frac {{\mathrm e}^{-2 i a} \expIntegral \left (1, -2 i b \,x^{2}\right )}{8}-\frac {{\mathrm e}^{2 i a} \expIntegral \left (1, -2 i b \,x^{2}\right )}{8}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x^2+a)^2/x,x,method=_RETURNVERBOSE)

[Out]

1/2*ln(x)+1/8*I*exp(-2*I*a)*Pi*csgn(b*x^2)-1/4*I*exp(-2*I*a)*Si(2*b*x^2)-1/8*exp(-2*I*a)*Ei(1,-2*I*b*x^2)-1/8*
exp(2*I*a)*Ei(1,-2*I*b*x^2)

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Maxima [C] Result contains complex when optimal does not.
time = 0.37, size = 51, normalized size = 1.38 \begin {gather*} \frac {1}{8} \, {\left ({\rm Ei}\left (2 i \, b x^{2}\right ) + {\rm Ei}\left (-2 i \, b x^{2}\right )\right )} \cos \left (2 \, a\right ) + \frac {1}{8} \, {\left (i \, {\rm Ei}\left (2 i \, b x^{2}\right ) - i \, {\rm Ei}\left (-2 i \, b x^{2}\right )\right )} \sin \left (2 \, a\right ) + \frac {1}{2} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^2/x,x, algorithm="maxima")

[Out]

1/8*(Ei(2*I*b*x^2) + Ei(-2*I*b*x^2))*cos(2*a) + 1/8*(I*Ei(2*I*b*x^2) - I*Ei(-2*I*b*x^2))*sin(2*a) + 1/2*log(x)

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Fricas [A]
time = 0.37, size = 39, normalized size = 1.05 \begin {gather*} \frac {1}{8} \, {\left (\operatorname {Ci}\left (2 \, b x^{2}\right ) + \operatorname {Ci}\left (-2 \, b x^{2}\right )\right )} \cos \left (2 \, a\right ) - \frac {1}{4} \, \sin \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x^{2}\right ) + \frac {1}{2} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^2/x,x, algorithm="fricas")

[Out]

1/8*(cos_integral(2*b*x^2) + cos_integral(-2*b*x^2))*cos(2*a) - 1/4*sin(2*a)*sin_integral(2*b*x^2) + 1/2*log(x
)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\left (a + b x^{2} \right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x**2+a)**2/x,x)

[Out]

Integral(cos(a + b*x**2)**2/x, x)

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Giac [A]
time = 0.44, size = 35, normalized size = 0.95 \begin {gather*} \frac {1}{4} \, \cos \left (2 \, a\right ) \operatorname {Ci}\left (2 \, b x^{2}\right ) + \frac {1}{4} \, \sin \left (2 \, a\right ) \operatorname {Si}\left (-2 \, b x^{2}\right ) + \frac {1}{4} \, \log \left (b x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^2/x,x, algorithm="giac")

[Out]

1/4*cos(2*a)*cos_integral(2*b*x^2) + 1/4*sin(2*a)*sin_integral(-2*b*x^2) + 1/4*log(b*x^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\cos \left (b\,x^2+a\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x^2)^2/x,x)

[Out]

int(cos(a + b*x^2)^2/x, x)

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